Factor completely. $5x^2-20x+20=$
Explanation: First, we take a common factor of $5$. $5x^2-20x+20=5(x^2-4x+4)$ Now, let's factor $x^2-4x+4$. Both $x^2$ and $4$ are perfect squares, since $x^2=({x})^2$ and $4=({2})^2$. Additionally, $4x$ is twice the product of the roots of $x^2$ and $4$, since $4x=2({x})( 2)$. $x^2-4x+4 = ({x})^2-2({x})( 2)+({2})^2$ So we can use the square of a difference pattern to factor: ${a}^2 -2( a)( b)+ {b}^2 =({a}-{b})^2$ In this case, ${a}={x}$ and ${b}={2}$ : $ ({x})^2-2({x})( 2)+({2})^2 =({x}-{2})^2$ $\begin{aligned} 5x^2-20x+20&=5(x^2-4x+4) \\\\ &=5(x-2)^2 \end{aligned}$ In conclusion, the complete factorization is $5(x-2)^2$ Remember that you can always check your factorization by expanding it.